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How to calculate lateral velocity of a turning aircraft
An aircraft in a 45 degree banked turn has a horizontal lift force vector equal to the weight of the aircraft.  Although the Coefficient of Drag is much larger for the aircraft profile, it seems there must be a significant side drag force on the aircraft as a result of the lateral velocity created.
Using the ROT formula:  1091 × Tan bank angle/V knots
a 45 degree turn at 100 knots gives an ROT of around 11 degrees/sec
Sin 11 degrees x 100 is around 19 knots lateral velocity.
Is this an accurate number?  Must we consider this side force even if the plane is "coordinated" from the pilots perspective?
Because the nose is also rotating into the turn, this number may be less.  Is it more than 0?
Are these the vectors describing a coordinated turn?

“Side drag”?  What’s missing from established explanations about aircraft turning that prompts such new terminology?  (Or dare I ask…)

This is Physics 101.  The velocity vector is "straight ahead" and tangent to the circle. Period dot end.  True on a 2D chalkboard, and still true in a 3D atmosphere with bank angle and lift components and all the rest.

If velocity is not changing (in your defined Frame of Reference), then side force is zero. Problem is, What frame of reference are you measuring things in? The airmass? The Earth?, the Aircraft? In each of these, the measured velocity will be different, and to get the equations to balance you may need to add one or more fictitious forces, (or a "SideForce") to make it work out. Without specifying (and talking about) the frame of reference, everything else is ambiguous and undefined.

@RalphJ so what you're showing on your blackboard is sideways movement, driven by the wing.  Extra lift (therefor extra drag) is borne by the wing.  Agree the velocity vector is straight ahead (projecting from the board) and the fuse as aligned with the relative wind as much as possible.  Again thanks, before moving on to turning techniques specific for aircraft such as hang gliders.  Though there is a lot of sideways acceleration, constant re-alignment of the nose to the turn minimizes side velocity, like catching a 10 m/s$^2$ accelerating stone after a 2 inch drop.

An aircraft in a 45 degree banked turn has a horizontal lift force
vector equal to the weight of the aircraft. Although the Coefficient
of Drag is much larger for the aircraft profile, it seems there must
be a significant side drag force on the aircraft as a result of the
lateral velocity created.

A sideways force component does not imply a sideways velocity component. That is an Aristotelian way of thinking. Nowadays, we understand that a force component drives an acceleration component, not a velocity component. And turning flight is a form of acceleration, even though the magnitude of the total velocity vector may remain constant.  The aircraft is constantly accelerating toward the center of the turn.  But this does not imply that there is any velocity component towards the center of the turn.

How to calculate lateral velocity of a turning aircraft

In a "coordinated" turn1, the lateral velocity toward the center of the turn, in the airmass reference frame, is always zero, because the aircraft is always pointing in the same direction as it is moving through the air.2
The critique of the actual formulae presented in the question, will be left to other answers.
Footnotes:

It's important to note that turns tend not to be completely coordinated  unless the pilot (or the autopilot) is making a rudder input to accomplish this.  An aircraft has some tendency to sideslip-- the nose tends to point slightly toward the "outside" or "high side" of the turn.  This tendency is typically most pronounced as an aircraft is entering the turn, but is typically present to some degree even in a constant-banked, steady-state turn, especially at lower airspeeds, or more precisely, at lower "scale speed", which is inversely related to the time to cover one fuselage-length.  The reasons for this involve the difference in airspeed (and therefore drag) between the inboard and outboard wingtip, the effect of the curving flight path and relative wind on the vertical fin, etc, and are best addressed in detail elsewhere, but this effect does have a very significant effect on lateral stability.  It is the reason that geometries like dihedral, sweep, and the multiple different effects related to high-wing or low-wing geometry, do affect an aircraft's lateral stability, including an aircraft's rolling-in or rolling-out tendencies in a stabilized constant-banked turn, in a situation where the pilot is making roll inputs as needed with the ailerons, but is not using the rudder to ensure that the turn is completely coordinated.  Calculating this sideslip angle in any given situation is far beyond the scope of this answer!

It's pretty obvious that the question was intended to be about velocities in the airmass reference frame, or in no-wind situations.  Computing the lateral component of the groundspeed at any given instant is a solvable problem, but best addressed elsewhere.

Just want to be sure.  What you say makes sense, and this is why PK concluded has "super high wing" would do fine in a turn.

@RobertDiGiovanni -- added content in what is now the first footnote, may be helpful--

Multiple points made in this answer could be associated with links to other ASE answers where the point is developed in more detail-- don't have time to do a full search right now, stand by--

No velocity component towards the turn, got that, but in the context of an orbit the plane moves a certain distance forward and a certain distance to the side (while it rotates to maintain low drag).  It does not have a velocity component towards the center because the forward motion and inward acceleration keeps it at the same distance.  This is for academic pursuit, we all know coordinated turns are best, but what does a straight yaw string on a banked aircraft tell us?

@RobertDiGiovanni -- it's a very close approximation-- and as far I've seen in personal experience it's entirely true as far as I've been able to see w/ my own eyes-- that if circling with ball centered, then yaw string is centered.  Based mainly on observations in sailplanes possessing both instruments; obviously in single-engine planes with propeller in front, spinning propwash messes w/ yaw string at least a little bit.

@RobertDiGiovanni -- but to be 100% accurate you'd have to account for fact that rotational motion of plane means yaw string at nose does not read exactly the same as yaw string further aft, say at CG.  But these are tiny tiny effects, that I can't say I've ever been able to directly observe firsthand.

Yes the airplane will always be turning with the "weathervaning" tendency from the fin lagging slightly behind the lateral acceleration, so even a turn with no adverse yaw present will be slightly slipping absent a rudder input to help the fin (which is what makes dihedral effect work, and why the fin shouldn't be too large).  But I think the slip is too small to discern clearly from either the skid ball or the yaw string in regular flying, without having a calibrated instrument to do the measuring in perfectly still air.

It would seem the major drag penalty in our 45 degree turn would be 40% more induced drag to make the horizontal lift, which drives the lateral acceleration force. I guess all we can do to help is to stay coordinated.  Thanks again.

@JohnK -- re "lagging"-- in my conception, after the bank angle has been stabilized for more than a few seconds, the issue with the fin's effectiveness is not any kind of a "lag", but rather a steady-state effect effect due to the fact that the localized relative wind is actually curved to conform to the curving flight path (at least in the simple case where the aircraft's sideslip angle and yaw rotation rate have settled down to constant values).  Curving relative wind "pushes" the fin inward as hinted at here -- https://www.av8n.com/how/htm/yaw.html#sec-long-tail-slip

Re "the localized relative wind is actually curved to conform to the curving flight path" -- this being due to the fact that the local relative wind at any given point can be expressed as the sum of the relative wind due to the aircraft's forward velocity measured at some arbitrary reference point (such as the CG), and additional changes in local relative wind velocity due to the aircraft's rotation rate in relation to said reference point--

Key words of wisdom from that except -- "This slip sounds like a bad thing, but in fact it can be put to good use; without it there would be no roll-wise stability, for reasons discussed in section 9.3. Remember: an inadvertent turn will be a slipping turn."

Again, before talking about or performing an analysis of the forces, velocities, and accelerations involved in any scenario, it should be specified and defined as to what frame of reference the comments, discussion, analysis is going to be done in. Indeed, the choice as to what frame of reference we will do the analysis in is often the most important decision affecting the ease of the analysis, and the clarity of the discussion around it.

Easy calculation: lateral VELOCITY is ZERO in a coordinated level turn, using the air-mass as the frame of reference.
Lateral ACCELERATION is non-zero, and is moving the velocity vector at the same rate that the heading is changing. The velocity vector is, in a coordinated turn, aligned with the nose of the aircraft, so lateral velocity can only be zero.
The uninteresting exception to this occurs when using a ground-based frame of reference and wind exists. At that point, you would have lateral velocity to the extent that the wind gives you some crosswind component. But that's the case also in straight & level flight, and is unrelated to all dynamics of the aircraft & its turn.

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How close is the air resistance or drag force to being proportional to the airspeed squared?
How is the total air resistance force exerted upon a body (aircraft or car) related to its speed relative to air at far away? A simple Newton's force momentum analysis suggests that the force $F \propto v^2$ where $v$ is the speed of the body. I am sure it is not exactly speed squared but how close is it for a car?

In the aircraft case, are you assuming a fixed angle-of-attack (such as would be the case if we mounted the aircraft in a wind tunnel and varied the wind speed), or are you assuming the angle-of-attack is varied as needed to maintain level flight with lift equal to weight?  It will make all the difference as to the correct answer; the one answer given so far appears to be assuming the latter case.  Of course you can see why an aircraft considered according to the latter assumption, would be governed by very different equations than a car where the weight is carried by the wheels.

@quietflyer: I did not consider the distinction. It would be very informative if you would write an answer elucidating the two scenarios in details.

Peter Kampf's answer assumed the latter case-- hence the statement "... induced drag ... decreases with speed squared."  and that would be the normal assumption when talking about an aircraft in the context of actual flight.  Just understand that the same conclusions wouldn't be valid for talking about what happens when you mount an aircraft at a fixed angle-of-attack in a wind tunnel and vary the wind speed.  No component of drag will decrease as speed increases in the fixed-in-wind-tunnel case, but lift won't stay equal to weight either.

Subsonic aircraft drag is composed of two major components:

Pressure drag. Most of this is induced drag, but also some is from viscous effects which result in less pressure on the rear sections of aircraft.
Friction drag from boundary layer effects.

At high speed, friction can be the dominant drag contributor and here we have a less than quadratic dependency because with increasing speed the influence of friction is diminished. This is expressed by the Reynolds number and for first order approximations the friction drag coefficient changes in proportion to $Re^{-0.2}$. Since Re increases linearly with speed, the friction drag force is approximately proportional to $v^{1.8}$ at high speed.
At low speed induced drag is dominant and this decreases with speed squared. Of course, there is still a contribution from separation (which is a major problem at low speed), so the proportionality is smaller. Still, at low speed aircraft drag drops over speed.
With cars things are simpler. Here we have a rather blunt body with lots of separation at the backwards-facing surface. The suction on that separated flow is the dominant contributor to a car's aerodynamic drag and is much less influenced by Reynolds number effects. However, a small decrease in the friction drag coefficient over speed on the surfaces with attached flow remains, so the proportionality is a whisker below 2. For engineering, using $v^2$ is good enough, however.

Thank you. With regard to cars, could you please give a reference to the statement "a small decrease in the friction drag coefficient over speed on the surfaces with attached flow remains, so the proportionality is a whisker below $2$"?

@Hans: Some of the total drag is of course friction, and here things are like on aircraft with a bit less than $v^2$ proportionality. Only that pressure drag from separation is much bigger on cars as a proportion of total drag, so the influence from friction is smaller.

+1 and accepted. It would make it easier to read if you write out the explicit formulae taking into account your ponts 1. and 2. It took me a double take to realize by "the friction drag changes in proportion to Re$^{−0.2}$" you meant to say the frictional drag coefficient $c_f$ rather than the friction drag itself is proportional to Re$^{−0.2}$. More explicitly $D_f\propto c_fv^2\propto Re^{-0.2}v^2\propto v^{1.8}$ .

@Hans: Yes, you are correct. That should be the coefficient. Sorry, corrected!

This good answer might be further improved by explicitly pointing out that the angle-of-attack of the aircraft is not held constant as the airspeed varies, and that's the reason why the induced drag component decreases as airspeed increases.  So naturally, the equations involved in driving a car at varying speeds will be quite different than the equations involved in actual flight.  Likewise, mounting an aircraft at one fixed angle-of-attack in a wind tunnel and varying the windspeed will not tell us much about the drag force we'll experience in actual flight over a wide range of airspeeds.

@quietflyer: Yes, that is one way to illustrate the difference. However, it would also be possible to keep the AoA constant and change flap settings with speed. In the end, it is the need to produce a given amount of lift in varying speeds that makes aircraft different (then again, cars at high speed also can produce substantial lift …).

Just an aside about the car producing lift (or downforce for that matter) -- if we assume that the car's "angle-of-attack", and therefore lift coefficient, stays constant (maybe not a valid assumption due to springs in suspension etc), then the fact that the car is producing lift may not really complicate things very much.  Unlike the airplane case.

Would you mind writing out the explicit formulae for the total drag taking into account your points 1 and 2 so as to facilitate comprehension? Thank you.

@Hans There is no simple formula. Instead, drag is a bit like accounting where all contributors need to be added up without counting one twice. Next, you have an influence from surface imperfections down to the roughness of the paint job or details of the body contour like air intakes or bumps.

Keep in mind that a car always presents the same "angle-of-attack" to the airflow (ignoring crosswinds, and ignoring changes in the car's pitch attitude due to effects related to the stiffness of the suspension springs), and so does an aircraft rigidly mounted in a wind tunnel.  Thus the drag force can be expected to vary according to velocity squared in such cases, except for complications due to changes in Reynolds number, etc.
However, an aircraft in actual flight is constrained to have lift equal to weight (assuming the simple case of straight-line horizontal flight with no upward or downward tilt in the thrust vector), and thus must fly a higher angle-of-attack at low airspeed than at high airspeed.  Therefore the lift coefficient and drag coefficient vary dramatically throughout the flight envelope, and so total drag cannot be expected to vary according to airspeed squared or anything close to that.
Here is a link to an excellent on-line resource explaining how to calculate the drag created by an aircraft in flight -- https://aerotoolbox.net/drag-polar/ .  Scroll down to the interactive graph entitled "Variation in Aircraft Drag with Velocity" near the bottom of the webpage, specifically the third-to-the-last figure, with Velocity (kts) on the horizontal axis and Drag force (Newtons) on the vertical axis.  You'll see that the drag force is nowhere near proportional to the square of the velocity for this part of the flight envelope, due to the fact that the induced drag component is decreasing as the airspeed is decreasing.  At some vastly higher airspeed, where the induced drag component is trivial, this particular aircraft will experience a drag force that is closer to proportional to the square of the velocity (at least for subsonic airspeeds) than is shown in the part of the flight envelope illustrated here.
However, if the aircraft were intended to fly that fast most of the time, it would be a waste to design it to have such a large wing, due to the profile drag and skin friction drag created by such a large structure.  With a smaller wing, the aircraft would fly at a higher angle-of-attack at any given airspeed, and induced drag would no longer be trivial at some given higher airspeed where it was trivial with the larger wing.  In fact, the minimum total drag (and thus the minimum thrust requirement for horizontal flight) occurs at the airspeed where the induced drag contributes half the total drag, so there is no point in a designing an aircraft with an overly large wing in an effort to minimize induced drag at the cost of increased profile drag and skin friction drag.  In short, in the part of the flight envelope where a given aircraft typically operates, induced drag is usually not trivial, and therefore total drag is nowhere near proportional to airspeed squared.

Yes as also alluded to by another comment under another answer, this answer doesn't consider changing flap settings, etc.  Even when you take those things into account, you still get a drag versus airspeed curve shaped more or less like the one illustrated in the graph referenced by the link given in this answer.

"In short, in the part of the flight envelope where a given aircraft typically operates, induced drag is usually not trivial, and therefore total drag is nowhere near proportional to airspeed squared."-- and likewise is nowhere near proportional to airspeed ^ 1.8.

+1. Very informative answer.

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Simultaneous yaw damping and sideslip control of aircraft with rudder?
Yaw damping is achieved by using the rudder to generate a yawing moment that opposes any yawing motion. Also, rudder is used to correct sideslip by applying a rudder input in the opposite direction of the sideslip.
My question is how to simultaneously fulfill yaw damping and zero side-slip/yaw acceleration requirements? More specifically, should both requirements be controlled independently and resultant rudder is the sum of two or yaw damping should be made inner loop to sideslip/yaw acceleration control?

It's all a single activity.  Yaw dampers provide a single function; provide an active automatic rudder input to keep the airplane aligned in the airflow when the airplane's static yaw stability, its "weathervaning" tendency provided by the vertical fin, is insufficient.
The system detects sideslip for any reason and applies rudder to take out the side slip, within its range of authority (usually around 1/3 of rudder travel). It's normally a separate always-on channel, not connected to the autopilot.
Think of it as a separate person you hired to sit with his own set of rudder pedals, and instructions to watch the slip/skid ball and do whatever it takes to keep the ball centered at all times.
This means that displacements in yaw from turbulence or gusts are automatically controlled, as well as adverse yaw induced by aileron displacement.
When an airplane turns, the turn is generated by the bank created by aileron, which causes sideslip toward the low wing. The vertical fin provides the yawing action to coordinate the turn; a big weathervane (as the bank steepens, more and more of the turn rate is coming from elevator, but for simplicity we'll assume very shallow turns where that is minimal). Rudder inputs are only required for turns because of adverse yaw from aileron displacement, or gusts that excite yawing movements.
On an airplane with a yaw damper, you put your feet on the floor after takeoff and never touch the rudder pedals during normal maneuvering in the air. The yaw damper does all the corrective inputs to keep the ball centered for you. Outside the take off and landing phases, you only need to get on the pedals for situations that require more input than the yaw damper has authority to correct. Namely, engine failures.

Sir, in case of yaw damping, we measure yaw rate and try to apply rudder so as to bring yaw rate to zero, while in case of sideslip control, we measure sideslip or lateral acceleration and try to bring it to zero by applying rudder. Are not these two different processes meaning performing yaw damping in this sense can't provide sideslip control automatically and vice versa. Kindly clarify.

Yaw rate, the speed of the rotational motion about the yaw axis, only comes into it in determining the Y/D's response characteristics, how fast it responds, how much input to make, etc. In the end it's doing its thing to eliminate lateral acceleration - slip/skid. You have a lateral acceleration, or you don't. If you are just looking at the skid ball, it's centered, or it's not and you're slipping or skidding, whatever you're doing at that moment. If it's off center, you'll apply rudder to center it. If the Y/D is engaged, it will do it for you.

the resultant rudder is the sum of the two.

If you want to fly a straight path, yes.
Sideslip is the result of an uncoordinated yaw.  A yaw damper is required when the aircraft's natural damping ability is insufficient.
Exacerbating factors are high altitudes, high weight to surface area ratio, and swept wings.  A paper airplane does not need a yaw damper, a 747 flying at high altitudes does.
In design the vertical stabilizer and fuselage surface area behind the center of gravity control yaw.  This is known as longitudinal stability.
The need to actively apply rudder to eliminate unwanted sideslip and/or damp yawing oscillations indicates the aircraft is more unstable in the regime it is flying.
Interestingly, in the design of very large aircraft, a yaw damping system may prove more economical than an extremely large tail fin.  However, larger fins are virtually foolproof safety insurance.

More specifically, should both requirements be controlled
independently and resultant rudder is the sum of two or yaw damping
should be made inner loop to sideslip/yaw acceleration control?

Except when operating near the stall -- and striving to avoid a spin-- applying rudder inputs as needed to keep the ball centered will always give satisfactory results.  Assuming that the pilot (or the automatic flight control system) is able to keep the ball truly centered, this statement holds true even in parts of the flight envelope where an aircraft (typically a swept-wing aircraft) has a self-sustaining (undamped) "Dutch roll" yaw-roll oscillation mode.

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Is drift and sideslip angle the same in the accompanied figure?
Consider the following figure:

Initially aircraft is moving along dotted line. But due to wind, aircraft ground track is changed to solid line but body still pointing parallel to dotted line. I've the following questions:
(1) What is sideslip angle in this case?
(2) What is direction of relative wind?

Both your suggested phrases are depicted visually in the diagram.  I don't think they need a written explanation to clarify.  A picture is worth a thousand words, right?!  ;)

Relative to what?  To the plane it's directly in the line of flight.  To the ground it's the vector sum of plane velocity and the (plane and) airmass velocity.  Just draw the vectors. So the plane heading will be equal to the drift angle, but in the opposite direction, in this case around 070 and the drift will put one on course at 090.

The direction the pilot wishes to fly (desired course) really isn't relevant to anything. Still, there may be enough information here to answer all the questions.  The question that should have been put to the O.P. before anyone answered, was "do you intend the arrows to be vectors, i.e. does the length of each arrow correspond to the magnitude of the variable involved (groundspeed, wind)?"  (Using "vector" in the physics/ engineering sense, not the ATC sense. which is just a direction to steer.)

The answer to the title question is no, sideslip angle and drift angle are not the same thing.  (in this figure, or any other...)
What the diagram shows is drift angle, and drift angle only.  If you are moving through a uniform airmass that is itself moving relative to the ground, your path across the ground (ground track, or course) will be affected by the movement of the air.  The difference between your heading (simply the direction the aircraft is pointing) and your actual ground track is the drift angle.  That is the teaching point, and purpose of this figure.
Regarding the numbered questions:

There is not enough information in the diagram to determine this.  Sideslip angle is the lateral angle between the relative wind and the longitudinal axis of the aircraft.  If the aircraft is in balanced flight, with the balance ball centered in the slip indicator, the sideslip angle will be zero.  It doesn’t matter what your heading is, it doesn’t matter which direction the wind is moving, or whether the aircraft may be drifting from its intended course.

The direction of the relative wind is always opposite to the direction the aircraft is traveling through the airmass.  In this example, presuming balanced flight, the relative wind is coming directly from the East.  If the aircraft was unbalanced, (balance ball out to one side or the other) there would be a sideslip, and relative wind would be striking against the side of the fuselage instead of head-on.  That angle is the sideslip angle.

There are different and better illustrations that show sideslip angle if you need more help understanding what it is.

Thanks for the answer. Regarding "There are different and better illustrations that show sideslip angle if you need more help understanding what it is", kindly illustrate for my better understanding of sideslip and relative wind.

Just enter “side slip angle” in Google and click on the “images” tab.

Re "The answer to the title question is no, sideslip angle and drift angle are not the same thing. (in this figure, or any other...)" -- Elsewhere you've defined the drift angle as the angle between the aircraft heading and the ground track (see https://aviation.stackexchange.com/questions/95979/calculate-measure-sideslip-angle-from-heading-track-and-drift-angle/95986#95986).  So, by this definition, isn't the drift angle the same as the sideslip angle, any time an aircraft is sideslipping in no-wind conditions?

@quietflyer, technically yes, but since the goal in most instances is to fly with zero sideslip, and since the illustration doesn't depict a sideslip angle, and because in cross country flight planning 101 and E6B training we aren't taught to factor in a planned  side slip angle and adjust heading to achieve it, my answer is correct for all but those who choose to mitigate the effects of wind with rudder alone.  Although the definition would then be incorrect because the aircraft really isn't drifting, it's just misaligned.

I'd submit that if the arrows are intended to be vectors, i.e. if length of groundspeed (or ground track) and wind arrows are intended to be proportional to the groundspeed and wind speed, then there is enough information here to say whether the plane is sideslipping or not.  That's the basis of my answer.  (But unfortunately the O.P did not clarify whether that was his/her intent or not.)  Also, it appears most likely that the O.P. had some confusion about what sideslip means.

The relative wind (as felt by an imaginary weather vane on the aircraft) is coming from straight ahead. The side slip angle is thus 0.

Since we all seem to be going "enhanced response" on this answer, let's start by interpreting the picture in the question:  an aircraft attempting to fly 090 but is being blown off course.

with these parameters:

 airspeed:  100 knots
 crosswind: 15 knots from the north
Drift angle:  arc tan (15/100) = 8.5 degrees
Groundspeed: cosine 8.5 = 100/x
  x = 101 knots
If the pilot wanted to fly 090, then the heading would be 081.5 to correct for the crosswind.
Ground speed would be cosine 8.5 = x/100 (because heading is now the hypotenuse) = 99 knots.
One can see that correcting for a 90 degree cross wind does indeed slow groundspeed slightly.

But this doesn’t answer the questions.   OP doesn’t ask about GS.

@Jim relative to the ground point, the plane is "sideslipping", relative to the airmass, it isn't.  More reading here.

He’s crabbing into the wind, but sideslip is zero (for his assumed coordinated flight).  But my comment was because you go to great lengths to show that ground speed is decreased when crabbing into the wind.  But that proof isn’t what OP was asking about.

@Jim acknowledge that point.  I felt it kind of helps understand the geometry to discuss airspeed and groundspeed for both the drift and to hold 090.

Question is (from one perspective), flawed, or at best incomplete. You are dealing with two different frames of reference. 1) the frame of reference of the atmosphere, and 2) the frame of reference of the earth. Every variable or measurement must be identified and labeled to indicate which frame it is being evaluated in.
for the sake of argument assume the wind is 5 knots and the aircraft speed is 100 knots.
First, the word "Track" implies the earth frame of reference. but the word Heading implies only the orientation of the aircraft. But let's assume Track can be measured in either frame...
In the Atmosphere, aircraft Heading is 090 Due East, Track (within the atmosphere), is also 090 (due East), Sideslip (difference between Fuselage reference line and relative wind), is Zero. airspeed will not change (100 knots). "Course" is meaningless as it is a concept relative to the earth frame of reference, not the atmosphere. It is actually somewhat the same as "Track".
Relative to the Earth frame of reference (in your diagram):
aircraft Heading is still 090, Track (relative to the ground now), is south of due East (about 095), Sideslip (difference between Fuselage reference line and relative wind), is STILL Zero. airspeed is a concept relative to the atmosphere, so it is not directly measurable in the earth frame of reference. Ground speed on the other hand, will be higher (due to vectoral addition of small tailwind component -maybe 103-105 knots). "Course" is just "track" in the frame of reference (across the ground), so it will be to the south a bit (about 95 degrees).
By the way, in reality, if you actually did this (Take off on a runway aligned due East with a left cross wind), what would happen is this:
First of all, to keep the aircraft tracking straight down runway centerline, prior to liftoff, you would need to add DOWNWIND (right) rudder to counteract the wind-milling effect of the crosswind on the vertical stabilizer. Without it the aircraft would turn upwind - into the wind. You would also need to add upwind (left) aileron to keep the upwind wing from rising and to prevent the roll away from the wing that the rudder would otherwise cause.
As the aircraft lifted off (as the wheels lost contact with the runway), if you kept those cross-controls applied, the aircraft would be flying on a due easterly ground track, in a sideslip, Tracking straight down extended runway centerline, with a true airspeed velocity vector upwind (slightly north) of the runway centerline (about 085-090 degrees). (i.e., it's velocity vector IN THE MOVING ATMOSPHERE Frame). It would be moving slightly north of east through the atmosphere, as the atmosphere is moving south at the same rate, vectorially cancelling each other out to produce a ground track aligned with the runway (due east).
If at this point you neutralize the controls, what will happen is that aircraft will rapidly weathervane into crosswind, removing all sideslip. It will now be point "upwind" (080-085?), but will still be tracking due east along extended runway centerline.
Physics cannot be ignored, nothing (including aircraft) can change it's velocity without a force acting on it. An aircraft flying in a the air does not feel ANY force from "Wind". "Wind" is just the mathematical representation of the motion (velocity) of the entire atmosphere relative to the motion (velocity) of the earth (or the aircraft carrier, or the car you're riding in, or the train you're on, etc. etc. The only force an aircraft feels is from the "relative" wind, the motion of the air across the aircraft skin due to it's motion THROUGH THE AIR,  not across the ground. (and the thrust of the engines and it's weight).

Your use of heading and track is somewhat misleading. Track is usually a term reserved for the Earth frame. (Magnetic) Heading is defined w.r.t. the Earth's magnetic field, so it does not change between atmosphere and Earth frame of reference.

@Bianfable, This was deliberate. This implied meaning of track, mixed with assumptions about heading and ignoring difference between two frames of reference is the cause of the confusion.

OK, I think with the last edit it should be clear ;)

The question is not flawed, but there are two flaws in this answer: Heading is the same for either frame of reference, and 5 knots direct crosswind would not net you 105 over the ground if you were at 100Kias.  Also, crosswind takeoff technique wasn't really a part of the question.  Downvoted for these reasons.

@Michael, Perhaps "Flawed" is inappropriate or wrong word to use. I have edited it to attempt to be more clear. - but actual numbers are clearly NOT intended to be perfectly accurate. Adding trigonometric formula or extended decimal values would add nothing to intended meaning.

I wouldn’t expect any trig, but for 90 degree increment wind effects the basic math is straightforward enough.  (I.e.  at 100 kias with 5 knots of wind your GS is:  95 with a headwind, 105 with a tail wind, 100 for a direct crosswind.)  You added the crosswind as if it is a tailwind.

To address your issue I have edited answer to removed specific values... Probably clearer this way anyway... Thanks!

Well, I would like to say that's better, (and you did fix the discrepant heading in the earth reference...) but you are on strike two regarding the groundspeed.  A direct 90 degree crosswind has a tailwind component that is effectively zero.  The ground speed is 100, not 103-105.  Don't post it here, but go ahead and do the trig if it helps...  ;)

Ok, it isn't trig, instead we turn to Mr Pythagoreus.  I was curious, so I worked it out:  Groundspeed would be 100.1 knots if you made no correction and flew the hypotenuse.  Not even noticeable.  Nothing any pilot would ever care about.  Even the crab angle would probably be insignificant.  I won't bother figuring that one out.

@Michael,  If the crosswind remains 90 degrees to the ground speed vector you would be correct, but the higher the crosswind the more the atmosphere is moving sideways, and, as the true airspeed will not change, the vector sum of the two will increase. Just remember a 3, 4, 5 right triangle.... The hypotenuse measures 5, even though the other two sides (3 & 4) are indeed at right angles. In this case the crosswind remains at 90 degrees to the True Airspeed, not to the resultant ground speed vector.

Your comment "A direct 90 degree crosswind has a tailwind component that is effectively zero.", serendipitously reinforces my original point - the confusion of frames of frames of reference. It only has a zero tailwind component in the Atmosphere frame. In the Earth frame it has a tailwind component equal to the crosswind times the sin(drift Angle). (Imagine a 200 knot crosswind - surely the ground speed will increase, right?) You might also be thinking of tailwind in the Aircraft frame, where it is zero as there is no such thing as a cross-wind in the aircraft frame.. (only relative wind.)

I took this to "discussion" when prompted, but I don't see a link here for anyone interested in a rebuttal...

Nothing major to rebut/discuss anymore, as I see it...  Looks like we're more or les on the same page, picking at nits !

Except the nit is 30-50 times greater than actual...  ;)

Looks like the word "windmilling" was inadvertently substituted for "weathervaning" --

Disclaimer: since flying crosswise to the external, meteorological wind creates no inherent tendency for an aircraft to sideslip, as far as practical piloting is concerned we normally assume sideslip is zero in wind correction problems, in which case we could assume that the arrow labeled "desired course" is also the "airspeed" vector, simply because it is aligned with the aircraft heading. Therefore answering the particular questions asked here, in terms of the particular "givens" presented here (which do not seem to include an assumption of zero sideslip), is a purely theoretical exercise, with no real practical value.
This answer assumes that the all the arrows in the figure are intended to represent velocity vectors, whose length accurately corresponds to the speed (magnitude) of each of the three variables involved.1
The intention of the diagram appears to be that the "track", "wind", and "desired course" arrows could be re-arranged into a closed triangle.2,3  This means that the "track" (i.e. ground track or groundspeed) vector is exactly equal to the vector sum of the "desired course" arrow and the "wind" arrow.
This is a crucial piece of information.  It tells us that the "desired course" arrow represents not only the direction that the pilot wishes to track over the ground, but also the direction and speed that the aircraft is actually moving relative to the airmass.
In other words, the "desired course" arrow is also the "airspeed" vector.4
Therefore we can see that sideslip is zero-- the aircraft is exactly aligned with the flight path through the airmass.
The relative wind vector is always exactly equal in magnitude, and opposite in direction, to the airspeed vector.  So the direction of the relative wind is from right to left in the diagram.  Or in terms of an azimuth, using the usual convention of naming winds in the direction from which they blow, the relative wind direction is 090.

Is drift and sideslip angle the same in the accompanied figure?

No, the sideslip angle is zero, and the drift angle is not zero.
Note that since the final groundspeed vector is a "given", and the wind direction and speed is a "given", and the aircraft heading is a "given", and the final airspeed vector (and thus the final sideslip angle) can be computed, this is not really a question about what happens when an aircraft flies into a new wind field and the pilot makes no control inputs to try to maintain the original heading, zero-sideslip condition, etc. It's already "locked in" to the question that those control inputs were made, if needed. A question about whether any particular control inputs would be needed to maintain these conditions, either as the aircraft transitioned into the new wind field, or after the aircraft had come into equilibrium in the new wind field, would have to be phrased differently.
Footnotes:

If this is not actually the intention, then the question contains insufficient information to answer most of the questions, unless the reader is supposed to make some assumptions about what control inputs the pilot is or is not making, etc.

If this is really the intention, it would be a kindness to the reader for the illustrator to actually draw the arrows in the form of a closed triangle, so that the reader would not have to guess as to whether the arrows really all match up in this way or not.

A very close examination shows that the "wind" arrow is slightly too long to fit in a closed triangle with the other two arrows.  This answer assumes that this is not intentional.  If it were intentional, the "airspeed" vector would be slightly longer than, and aimed slightly counter-clockwise of, the "desired course" vector.  That would change the answers to questions 1) and 2), but not the answer to the title question.  Answering questions 1) and 2) for this alternative case is still entirely possible, and will be left as an exercise to the reader!  Note that this alternative scenario is only compatible with the "drift angle" label on the diagram, if we define the "drift angle" as the difference between the aircraft heading and the ground track, not the difference between the direction of the flight path through the airmass and the ground track.  Sideslip is almost always assumed to be zero in wind correction problems, but in cases where sideslip is not zero, there is arguably some ambiguity around the exact definition of the "drift angle".  For more on this, see this related ASE answer.

In addition to the argument based on the fact that the arrows appear to be intended to fit together in a closed triangle, we're essentially told in the first sentence under the figure that the "desired course" arrow also represents the airspeed vector before entering the new wind field, and we're not told of any changes in airspeed, power setting, or climb/sink rate, and we're told that the aircraft heading remains constant, and therefore it's only logical that the "desired course" arrow also represents the airspeed vector after coming into equilibrium in the new wind field.  But this line of logic is weakened by the fact that the question seems to entertain the possibility that the aircraft may be flying in a slipping condition after coming into equilibrium in the new wind field.  Therefore in the context of this question, the best way to find the flight path relative to the airmass is to do the vector subtraction of the wind vector from the ground track vector. A better presentation of the diagram would explicitly include the airspeed vector, i.e. the vector representing the flight path relative to the airmass-- but in labelling the diagram in such a way, the questioner would probably have realized the correct answer to all of his or her questions, before even posting!


text:
Will reducing the electricity power in an aircraft save fuel?
I was wondering if I reduce the power consumed by the aircraft, would the generator produce less power and if that would save some fuel.

Since all the energy available comes from the fuel, reducing the electrical load will save fuel. However, since you can't touch the aircraft systems, that leaves the passenger cabin. Forcing passengers to sit in the dark, eating cold food, with no in-flight entertainment, USB chargers or WiFi will likely cost more in lost revenue than would be saved in fuel.

What type of aircraft?

Yes, in theory, but you would have to do a total energy analysis to decide if it's worth it.  On the surface of it, generators consume fuel not being made into thrust. Watts being made to power things is watts not going out the tailpipe.
A good example of the potential benefit is the switch to LED cabin lighting. A 40w incandescent light is more or less a 35w heater that makes a bit of light.  LED lights drop the power consumption by 80-90%, although if you need heat in the same space, you have to replace the heat being made by the lights, so if you did an overall energy balance analysis in that case, you might find the actual saving is negligible for the period when you need to heat the cabin in cruise at altitude (because the reduction in heat from the lights has to be replaced by more bleed air demand).
If you need cooling, it's the opposite because the heat from incandescent lights is working against you, and it's a huge benefit to get rid of the heat produced by the lights, which reduces cooling demand.
So overall, there is probably some fuel burn improvement over time switching to LEDs, very small, but something (say, 200 30w lights consuming 6kW, replaced by 200 5w LEDs consuming 1kW, a 6hp saving, maybe 4lbs/hr - not nothing, but, but as I said you'd have to take more bleed when heating, offsetting a lot of that improvement, and you'd probably have to spread it out over quite a while to get serious money out of it).
Most airlines that switch to LEDs do it mostly for reliability reasons, where labor hours savings and parts costs make a stronger financial business case than just a few pounds of fuel spread out over time, although the fuel-saving does add to the total.
Another area would be advances in avionics that reduce power consumption (you can just imagine the drop in energy demands when avionics went from vacuum tubes to solid-state in the 50s).  Even there though, it's the same issue with heat that is not really wasted when the cabin requires more heating than cooling, that steals some of the benefits you thought you were getting.
Reducing electrical demand with efficiency improvements also helps with weight savings (smaller, lighter generators).
On the other hand.... the 787 blows all that out of the water by going to the other extreme, making all sorts of services electrical, requiring really massive generators. I'm not sure that there is a net energy benefit to using electricity to replace bleed for anti-icing for example, and I think the main benefit is reliability (bleed air systems are a massive maintenance and reliability hassle).

Note that the air often does require cooling even in cruise. Compressing the air to maintain cabin pressure is roughly adiabatic, so it also heats the air, and because the atmospheric lapse rate also corresponds to adiabatic expansion, you'll end up with approximately the same temperature as at 8000 ft. If it's hot on the ground, it may easily be more than you want in the cabin.

… but then, note that while heating the air does use some extra bleed air, and therefore energy from fuel, cooling it does not, because that is done by simply running it through heat exchanger with the ambient air (well, may be a little; I am not sure what effect it has on the pressures in the air cycle machine).

Well I think when you take in the interior volume that has to be heated and consider the high velocity -50C flow outside, with maybe 1/4" of insulation thickness, you have a net heating requirement and there is very little flow through the air cycle machines, only what is required to bring the raw bleed temp down to some value that is higher than ambient, and the heat exchangers will do most of that.  I've flown in airplanes at altitude with no interiors OR insulation above the cabin floor (corporate prior to completion center) and the cabin temperature at FL340 or so was about -20C.

On the other hand, the temp of the bleed is the temp of the bleed, and it is quite hot, and much of the excess heat is being dumped overboard in any case, with the flow being only what is needed to keep the pressure hull pumped up, so perhaps the difference is negligible.

The bleed air is very hot, but also higher pressure than the cabin, and when it expands to the cabin pressure, it cools again. So I don't think that much heat actually gets dumped.

When the air goes through the ACM it goes through several heat exchangers and/or precoolers on most systems.  They are bootstrap systems that remove some of the heat while the pressure is high with a precooler, boost the pressure and temp some more (ACM compressor) pass through more heat exchangers, and you end up with high pressure air that has had most of its heat removed.  When it finally goes through the ACM turbine, that's when the expansion happens and you get the final temp drop.  If you put all the air through an Air Cycle Machine, it's capable of dropping the output to -30 to -40C.

But some of the air bypasses the ACM and stays hot, and is blended with the cooled output to get the final temp going to the cabin.  The controller juggles this balance of direct hot air and cooled air to get the cabin duct temps it wants.

I am not sure how much it manages to expand the bleed freely rather than adiabatically to keep it hot, but I suppose it can do it enough to be useful, in which case there is probably indeed little difference.

Thank you all for the comments. When I asked this question, I was thinking about the retrofit of cabin ilumination indeed. I'm very interested in how you estimated the 4lbs/hr. Could you give me an exemple in how did you get to that number, utilizing an common aircraft like 737 or A320. Thanks in advance.

I just converted wattage to hp (746 w per hp), and used a number typical for the fuel burn of turboprops, which is in the region of .6 lbs/hp/hr. That would be a number you might get w a Q400 Dash 8 if it had that many lights. For a turbo fan you'd have to do a conversion using thrust to watts. You'd have to factor in the losses from the power conversion and transmission losses, maybe 5 or 10% or something (I have no idea exactly what), which might reduce the benefit a little. The CRJ "Next Gen"s went to LED lighting and the power drop was significant, but the main attraction was reliability.


text:
Why does a banked aircraft create sideslip?
Below is an excerpt from Raymer (Aircraft Design) where he explains the dihedral effect. He states that a banked aircraft (due to say a roll disturbance) will have a sideslip which leads to a restoring moment back to equilibrium. What is the physical explanation for why there is a sideslip?

The side slip happens because the lift vector is tilted.  Think of a hovering helicopter rotor tilting; the machine drifts to the low side of the rotor disc because the thrust vector of the rotor is tilted.  The wings of the plane are just making thrust, but moving forward at the same time.  Tilt the thrust line by banking, and a lateral thrust component emerges and the body moves laterally in response (the helicopter in forward flight is doing the same thing - you move the cyclic to tilt the rotor, and the machine moves laterally while going forward because the rotor's thrust line was tilted).
The introduction of the lateral thrust component means a reduction of the vertical thrust component, and unless total lift is increased, the body descends as it's moving sideways, so you have to raise the nose in the turn (or pull more collective pitch, or tilt the rotor back slightly, in the helicopter).  And since the lateral movement created by tilting the lift vector is occurring at the same time as the body is moving forward, the movement describes an arc.  A turn.
The nose follows the turn because the big weathervane at the back, the vertical fin, sees a lateral angle of attack resulting from the sideslip induced by the tilting of the lift vector, and it seeks to keep the fin's lateral AOA close to zero and so the tail follows more or less in line with the arcing path of the turn. And this is where we get to dihedral effect and sideslip.
You don't want the fin to do this without a little bit of lag.  This is the rub; for dihedral effect to work, some sideslip must be present to create the differential lift that creates the restoring or righting force.  So the fin needs to be sized to be effective at passively keeping the tail aligned in the airflow while the airplane moves laterally as result of being banked, but not so big as to be instantaneous at doing it; it needs to tolerate a bit of sideslip before it starts to work, so that dihedral effect can do its thing when small disturbances occur.
If you make the vertical fin too big, a bank induced by a disturbance is followed instantly by a yaw created by the incipient sideslip, to keep the tail aligned in the airstream, and nearly no sideslip occurs.  Plus the yawing movement means the outside wing is moving faster, tending to increase the bank more.  This airplane will tend to stay at its bank angle, or increase its bank angle and want to spiral.
Fin sizing has to find a sweet spot where just enough sideslip is tolerated to make dihedral effect work, but not too much.  Too big, and the airplane wanders off into a turn with the slightest disturbance because any amount of induced bank from a bump is following by immediate weathervaning toward the low wing. Too small, and lateral stability is good, but the nose hunts and wanders in bumps.
My own airplane is a homebuilt Pazmany PL-2, and it's a little under-finned, which combined with the inertial mass of fuel in the wingtips, makes the nose wander more in bumps than a lot of airplanes.  It has very little spiral tendency however and you can bank it 15 degrees in smooth air and leave it (too much bank for dihedral effect to work), and it will just stay like that, and a spiral dive takes long time to develop.

The act of rolling the aircraft creates adverse yaw, which generates sideslip. This is a transient effect which is explained in this answer.
In a steady coordinated turn (ball centered), however, sideslip will also be present. This is a steady-state effect. There are a few contributing factors:

There is steady-state non-zero rudder, which is necessary to zero out the yaw damping moment ($C_{n_r}$), yaw-roll damping moment ($C_{n_p}$) and adverse aileron moment ($C_{n_{\delta_a}}$). This rudder deflection creates an aerodynamic side force ($C_{y_{\delta_r}}$).
The presence of steady-state non-zero body yaw and roll rate induces side force ($C_{y_p}$ and $C_{y_r}$).

In a ball centered turn, the sum of aerodynamic side force is zero, which physically demands a non-zero sideslip to generate the necessary aerodynamic side force to cancel.
Together:
$$C_{y_\beta}\beta+C_{y_{\delta_r}}\delta_r=C_{y_p}\theta\frac{\dot{\psi}b}{2V}-C_{y_r}\cos{\phi}\frac{\dot{\psi}b}{2V}$$
This is with all engines operating, so no thrust asymmetry is present. And this steady-state sideslip is small for ball-centered turn.

Incorrect explanations abound for why an aircraft tends to sideslip when banked, in the absence of any "coordinating" rudder input.  Often the fact that the weight or gravity vector has a component pointing towards one wingtip is invoked.  See for example the diagram in Martin Simons' "Model Airplane Design".
The truth is that this situation exists in a coordinated turn as well as in a sideslip, and so this cannot be the reason for the sideslip. According to Newton's principles, a steady sideways force tends to drive a curvature in a trajectory, not a linear sideways motion.  It is not correct to say that the weight or gravity vector tends to cause the aircraft to slide toward the low wingtip whenever the aircraft is banked, just because the weight or gravity vector contains a component that points toward the low wingtip.
Here is one reason that an aircraft tends to sideslip when banked-- banking tends to cause a turn (a curvature in the flight path), and the resulting curvature in the flight path "bends" the airflow or relative wind.  In other words, as the aircraft starts to yaw in response to the changing direction of travel, different parts of the aircraft are moving through the airmass in different directions at different speeds at any given instant in time, and so different parts of the aircraft experience different velocities (speed and direction) in the local relative wind, which can be described as a "curve" or "bend" in the direction of the relative wind.  In the absence of any rudder input, the vertical tail experiences a relative wind component that tends to oppose the turn and slow the yaw rotation rate.  This tends to cause a sideslip.
If this is unclear, consider a pinwheel, or an aircraft in a flat spin.  As the body rotates, clearly, different parts of the body are moving through the airmass in different directions at any instant in time.
This phenomenon can also be described as "yaw damping".
This is also the phenomenon that is described as the "long-tail slip" effect in Section 8.10 of John S. Denker's "See How It Flies" website.
If the aircraft's fuselage could bend like a banana to conform to the curving path of the turn, then every part of the aircraft from the nose to the tail could be streamlined to the local direction of the curving relative wind, and this phenomenon would not occur.
Rotational inertia in the yaw axis also plays some small role in driving sideslip as the bank angle and turn rate increase.
Explanations that attempt to account for the relationship between banking and and sideslip, without discussing the curvature in the flight path that generally results from banking, are invariably incorrect.

Awesome answer! exactly correct! Proof is that if rudder was in front of the aircraft center of pressure (assuming the lateral instability could be controlled some other way), the induced sideslip would be in the opposite direction. In the F-4 on Check rides, we had to perform a maneuver where we pulled the nose up to 45-60 deg nose high, then unloaded to zero AOA, and rolled 360 deg as the flight path followed a ballistic arc, ending up pointed 45-60 deg down. As the aircraft rolled, to keep the ball centered, you had to input rudder towards the ground, with 60-90 degs bank angle lead.

This was necessary because as the flight path changed, continually bending towards the ground, the aircraft nose position was always lagging behind (upwards) from the the flight path, (which was bending downwards). So If you rolled left, then in the first 90 deg of roll, when the aircraft was in 90 deg left bank, the flight path was to the left of the nose, and you needed extra left rudder to "kick" it down, to the left, to align it with the moving flight path vector. Then in the last part, when in a right bank, you needed right rudder, to kick the nose down to the right, for the same reason.


text:
Do horizontal stabiliser force direction change with CG variation for a given aircraft?
I am wondering about the direction in which horizontal stabilizer provide force. I know that it can provide either upforce or downforce. But I am wondering whether is it determined for a particular airplane, I mean it is established for example that B737 tail will provide only a downforce and CG limits are set up in a way that when correctly loaded the tailplane will provide every single time a downforce or it depends from the CG location that tailplane will have to provide either a downforce or an upforce?

When I read the first sentence, I thought the rest of the question would be about horizontal component of the mentioned force.

You may find section 6.1.6 in "See How It Flies" to be instructive-- be sure and read the entire section to the end.  https://www.av8n.com/how/htm/aoastab.html#sec-pitch-equilibrium  .  See also section 6.1.1-- all the illustrations and explanations in section 6.1.1 are built around the idea of a LIFTING tail, further re-inforcing the idea that this is NOT incompatible with pitch stability -- https://www.av8n.com/how/htm/aoastab.html#sec-teeter

The ideal cruise condition, which minimises induced drag, is for the tail to exert zero lift. This avoids any wasteful tip vortices from the tail.
However for most planes the CG shifts during flight, if only because fuel is being used up. Also, trim conditions change with speed, rate of climb, altitude and a host of other variables during flight. The tail spends most of its time a little away from the ideal zero-lift condition, and it can go either way.
The "lifting tail" was much discussed in the early years of aviation. It has a destabilising effect which limits the stable CG range and every plane has a maximum safe level of lift from the tail while remaining stable. Nevertheless, away from that limit it contributes lift which helps to offload the wing and allows it to be that bit smaller.
According to Berriman; Aviation, Methuen, 1913.
"On some machines the tail planes are cambered like the wings and help to support the weight in flight. The angle of incidence of the tail is less than that of the wings, however, and the principle [of weathercock stability] apples equally to such cases." -- Page 69.
Berriman gives some supporting analysis in terms of CG and CP to explain what he summarises as weathercock stability. The subtleties of the lifting tail would be further elaborated in subsequent years.
The bigger the tail, the greater the CG range can be accommodated, and also the greater the temptation to offload the wing a little. The limit is the tandem-wing aircraft, where for stability the fore wing is loaded more heavily than the aft, but the aft still contributes substantially to the lift under all flight conditions. A tandem-wing modification of the Westland Lysander flew experimentally during WWII and test pilot Harald Penrose confirmed that its handling was more stable than the conventional version and had a greater CG range.
So ultimately, every plane has its own practical tail loading, which may be normally up, normally down, or variable around the zero point.
There is a myth that for a plane to be stable the tail must always exert a downforce. This seems to arise among students for two reasons. One is that, when the AoA increases, dynamically it must exert a larger relative change in downward moment about the aerodynamic center than the wing's upward moment. Diagrams showing this effect always have a large down-arrow at the tail and it is easy for the student to forget that this is a relative change not an absolute value. The other is that the wing works hardest at takeoff, when fully laden and at its slowest flying speed; when the tail pushes sharply down to rotate the nose up, the wing has to compensate for that as well, so this is the design condition for sizing the wing and everybody has to learn it. Put that first diagram alongside and you have a myth in the making. If you failed to understand any of this paragraph, then you too may fall for it!

There is a myth that for a plane to be stable the tail must always exert a downforce – finally, I'm not alone with that opinion here!

@PeterKämpf Don't worry, I'll continue sharing a link to the section about pitchwise equilibrium of "how it flies" in comments when needed.

@ManuH you have to do better than that av8n article written by who knows who.  Link me to a NASA or Boeing article or a university paper that describes what you guys are talking about.  I have had no luck.

So many folks miss the difference in horizontal stabilizer decalage and trim.  Guy correctly pointed out the tail force at optimal wing AOA (in cruise) is most efficient at zero, niether up or down.  It is the elevator/trim tab that accomodates the CG range.  One point I would question is (the horizontal stabilizer) "dynamicly, it must exert a larger downward moment" .  Actually, it is less, and, at higher AOA, it is supposed to help the wing push the nose down.  "Down force" tails do have a lower risk of stalling after the wing does, but poorly designed ones are not exempt.

Again.  Show. Me. An. Authoritative. Source..., like a NASA, Boeing, or university article that describes what you are saying, and I will eat my virtual shorts. Should be easy!  Should be everywhere no?

@RobertDiGiovanni You misquote me: I wrote ""dynamically, it must exert a larger relative change in downward moment" That change is of course the negative one you mention. The distinctions between absolute values, relative differences and rates of change need to be clearly borne in mind at all times!

@JohnK I gave an authoritative source in a comment on your answer. My recommendation is that you check out any basic degree-course-oriented text book on aerodynamics. If nothing else, it will tell you what to look for on the NASA web site.

Well, I posted part of an article from a Boeing website on my post.  Is that wrong?

Boeing are certainly wrong. The guy who wrote that never put their name to it. No peer review, no academic credentials, nada. It's sadly typical of the half-baked guff people often dish out to pilots who only need to fly it, not to understand its design. Never rely on company web sites! (and remember what Boeing never told the MAX pilots)

@JohnK The explanation of av8n is a single explanation working for all wings configuration (bi-wings like the prometeus, canard, classical downforce tail). One single explanation for many configurations is preferable to an ad-hoc explanation for each configuration. And authoritative sources such as the Pilot's Handbook of Aeronautical Knowledge can be incorrect

What about the Boeing article I linked?  Is that wrong?  The problem is av8n tries to lump dual lifting surfaces configs with conventional tail downforce configs when they are opposites.

Ahhhh... so av8n  right, Boeing wrong..  Cuz you say so.  Check.

@JohnK I also suggest you read chapter 3 of Caughey, David A. "Introduction to aircraft stability and control course notes for M&AE 5070." Sibley School of Mechanical & Aerospace Engineering Cornell University (2011).

@PeterKämpf -- re "There is a myth that for a plane to be stable the tail must always exert a downforce – finally, I'm not alone with that opinion here!" -- you've never been alone; during the few years that I've participated on ASE I've made multiple posts and offered multiple links in support of this--

In stabilized steady state flight, if the tail is providing lift UP, it means it is opposing a net pitch up moment (it's lifting to keep the nose from rising).  To have a net pitch up moment, the CG must be aft of the "mean center of lift" or Neutral Point, in order to create a nose up torque resultant from mass acting about NP.  This is a dynamically unstable condition and only FBW fighters can operate this way.
In any normal airplane, the CG must always be forward of the Neutral Point, this distance being called "positive static margin".  A positive static margin means there is always a net nose down pitching moment present and the tail has to provide down force to balance it out.  The only time the tail will lift up is during maneuvering cases where a stick forward control input is sufficient to pitch the airplane over faster than the natural pitching moment can do it.  Then the tail will be lifting, briefly.
If the tail was lifting in any normal steady state case, it would mean that that Static Margin is negative, and that to slow down, the trimable tail's incidence would have to move leading edge up and the elevator move down.  I am not aware of any aircraft in existence in this universe with a conventional tail where the elevator has to move down, or the stab has to move LE up, to slow down.
From Boeing:

1 Positive and Relaxed Longitudinal Static Stability
  In flight, the wings of a conventional airplane generate a nose down pitching moment. To balance this, a download is required on the tail. Airplanes loaded with an aft forward CG require less download on the tail.

Since download on the tail is negative lift, effectively increasing the weight of the airplane, the location of the CG affects the cruise performance of any airplane. Flying at an aft CG will reduce the download on the tail and improve cruise performance.
When airplanes are operated near the aft CG limit, download on the tail is minimized and angles of attack and drag are reduced. However, moving the CG aft reduces the longitudinal static stability of the airplane, something that all flight crews should be aware of (see figure 1 and "Static Longitudinal Stability and Speed Stability" below).
As airspeed varies from a trimmed condition, the column force required to maintain a new speed (without re-trimming) is a measure of static longitudinal stability. For any conventional airplane, the location of the CG has the strongest influence on static longitudinal stability. For a statically stable airplane the required column force, as speed varies from the trimmed condition, is less at an aft CG than it is at a forward CG. The minimum average gradient allowed by U.S. Federal Aviation Administration FAR Part 25 is one pound for each six knots. As the CG moves aft, it reaches a point where the stick force per knot drops to zero, then reverses. This location is called the neutral point. The difference between the actual CG location and the neutral point is called the static margin. With a CG forward of the neutral point, an airplane has a positive static margin and positive static longitudinal stability. At a CG aft of the neutral point, an airplane has a negative static margin, is statically unstable, and requires some form of augmentation to be flown with an acceptable workload.<

That old chestnut. It is true that the tail must provide a dynamic relative downforce when the AoA changes, but it can perfectly well be trimmed for a small net up-force in normal flight.

Sorry not buying it.  You are saying airplanes will fly with negative static margin.  Again, find me an AUTHORITATIVE source for this, NOT some other posting on here.  I have found nothing that says airplanes will fly with negative static margin, neither on line or in the books I have.  They all refer to a requirement for positive static margin, and to have positive static margin there has to be a ND pitching moment for the tail to oppose.  If the tail is lifting there is negative static margin.  If the tail was lifting, and speed decreased suddenly, the nose would pitch up.

Nope. Static margin and absolute tail downforce are not directly related, that is the cheap fallacy you have fallen for. Static margin is related to the change in up/downforce moment with AoA. If you cannot understand that, then you will not understand why you have been misled.

Fine. Refer me to an article outside of this forum, from say, NASA, that describes it.  Surely it's described as you say all over the place, right?  Should be easy.

JohnK Do your own Internet search. Easier for me to take the first standard textbook on my shelf. LJ Clancy; Aerodynamics, 1975, UK Pitman edn p.518. "the static margin is now defined directly as -(dCm/dCl)" where Cm is the pitching moment coefficient and CL is the lift coefficient. So its' all to do with the rates of change of coefficients, nothing to do with absolute tail load.

The rates of change of coefficients has to do with the tail's steeper lift slope, which is why they use thin airfoils.  The downforce of the tail declines more steeply then the lift of the main wing, allowing the nose to drop.  If the tail is lifting, the steeper decline in lift will make the nose pitch UP.  I wouldn't want to be in such an airplane.

@John K I submit a "tail lifter" can be stable within its design CG range.  Indeed, the XB-70 would be an extreme example of this.  A down lifting tail (with more pitch down authority) is good insurance with a swept wing design (because the wing tips go first in a stall). Uplifting tails are reputed to be more dangerous because most of their throw/trim is already used up.  But is one designs CG further back, an uplifting tail could be made to work.  Since they are less efficient, they are rare (essentially bi-planes).

@RobertDiGiovanni no you are describing tandem wings which is a different thing entirely.  A tandem wing, as with a canard, must have the forward wing with a steeper lift slope than the aft one. Coming up with an airfoil like that, which will give good flight characteristics, for a forward wing that is making the majority of the total lift is tough.  Rutan struggled with this until they came up with the U of Glasgow forward wing airfoil for the Vari Eze and follow-on designs which produces high lift plus desirable pitch and stall characteristics.

@JohnK. "If the tail is lifting, the steeper decline in lift..." has it totally back to front. If the tail is lifting, increase in AoA will increase its lift not reduce it. Really, with such basic howlers as this in your arguments, there is little point in my commenting further.

@GuyInchbald  remember the courtesy rules here.  Anyway, you are describing the tail's reaction to gusts changing AOA and applying a restorative tendency.  I'm talking about static stability and the ability of an airplane to maintain trim speed and positive stick forces.  If the tail is lifting in a steady state condition, the CG is aft of the NP, an unstable case, and stick forces reverse.  So the Boeing article I linked is wrong?

@JohnK I'm afraid Guy is right on this one

@JZYL ok so let's be clear.  You are saying that you can have the CG forward of the NP, the normal static margin, and with such arrangement, have a nose UP pitching moment, requiring the tail to lift UP to trim at the same time?

No, static margin is not related to tail lift direction. Your confusion stems from your belief that an aircraft pitches about the NP. That's not accurate: it pitches about the CG, always. NP is a stability point, not the point of rotation.

But in the static trimmed flight case, we aren't talking about rotations about an axis, we're are talking about static moments in balance about the net center of lift, the NP.  With the CG forward of the net center of lift, there is a ND pitching moment present, being the torque of the CG applying a gravitational force forward of the NP.  How can the tail be lifting up to maintain trim unless there is a NU pitching moment present, which would require CG aft of the NP? I have yet to see anybody here explain this.

In a static case, of course you can take the moments about NP. But NP isn't the CP of the wing-fuselage; so there's no reason to think that the moment from the wing-fuselage is zero. For the right CG and AOA, it can be positive at NP. In that case, the tail needs to provide up-lift to achieve static moment balance.

NP is the sum total of lifting moments so is the "net" center of lift taking into account wing, fuselage, nacelles, etc. (the whole problem with the MAX stemmed from the engines moving the NP forward).  In any case, if the tail is lifting, that means its AOA is below the zero lift line of the tail in order to produce up lift, which means the moment induced by air loads on the elevator surface is trailing edge UP (like an aileron), which is called control reversal, requiring forward stick pressure to maintain pitch, and to trim, the tab would have to work in the opposite direction.


text:
Can you help me identify a vintage airplane in a painting?

This is a recent photo of the king of Spain, Felipe VI, visiting Spanish troops stationed in Iraq. In the background there is a picture of an old plane, possibly with a propeller, that I suppose from WW2 era (more or less).
If that painting has been chosen to adorn a military installation located in Iraq and occupied by Spanish troops, I understand that it must be a type of historical significance for Spain or for Iraq.
Can anyone recognize it?

it could be a shared facility - the image may not necessarily be assumed to be entirely related to Spain

@NKCampbell ...as MikeSowsun's answer has shown... ;)

I was addressing the statement: "I understand that it must be a type of historical significance for Spain or for Iraq."

@NKCampbell Me too. I just found funny that I considered the possibility that it was an iconic plane for Spain and turned out to be a P-51 Mustang. :)

The aircraft in the painting (I’m not sure it is a painting?) is a North American P-51C Mustang in the Tuskegee Airmen colors of the U.S. Air Force in WW II.
I cropped the image above and then used TinEye Reverse Image Search to find the photo.
My search has expired on that website, but you can copy the “fuzzy” photo and try it yourself on the same site.
Search photo

Search Result

P-51C Mustang in Tuskegee Airmen (Minnesota wing of the Commemorative Air Force)

Well, a perfect answer. I used to wait a few hours before accepting them, but there's no doubt about it. Thanks a lot.

This is a fantastic answer, I just do not get the reverse search. You found the photo on which the painting was based? Or a (high resolution) photo of the painting? The planes are identical on the two first pictures (as far as I can tell from the blurry one). It must be a famous painting then, to have hi-res pictures of it (and not pictures which include it in the background, as in OP's case)

@WoJ, Who says that OP's picture is really a painting?  It could be a framed photograph.

@JPhi1618: I do not think that such photos were available at WW2 time (crisp, colors, ...)

@WoJ, its a recent photo of a classic, restored aircraft.  There are several operations around that keep WW2 aircraft maintained and operational for airshows and historic events.

@Ginasius better to wait a day.  SE folks are not paid staff, answer at leisure, and many of the smartest visit once a day.  You may have just missed them.  Although yes, this answer is unlikely to be bested!

@WoJ: Not only are the planes identical, but the clouds are as well.  Identical-looking aircraft from matching camera angles would be a possible coincidence, but matching clouds?  No way.

I voted this answer up not only due to the accurate answer, but because you show another tool how to search an image.

Your tineye says its expired

@JPhi1618 - ah right! I did not think about that possibility. Thanks, it all makes sense now.

My question would be➖why an American plane is in a painting or photo in that room in Iraq.

@WoJ Colour photography is much older than people realise: Kodak began selling Kodachrome colour film in 1935, and less convenient processes were commercially available even in 1907.

That room is in an American Military Base in Iraq. The Americans can hang anything they want on the walls.

@DavidRicherby: yes, I realize that. It is more the crispness of the photo and how colors are vivid which were not WW2-era like.


text:
Hypothetical - Basic Med & Insurance
I have a friend that needs some info...
He has a Basic Med certification in hand.  He takes a drug that is on the DNI list however it is admissible with a SI (SSRI).  His GP doctor signed his Basic Med form.  His GP doctor obviously knows the drug he is on and knows he is fully functional on the drug.  His GP doctor knows the drug is on the DNI list.
Given that background...
Would he have any possible insurance issues (renters or owners) if he were to ever get into an incident?  Could the insurance company deny a claim if he takes the drug even with a doctor's approval but WITHOUT the SI?
Thoughts?

Your friend is not legal to fly with BasicMed without having the SI. He still needs to get the SI. Not only would insurance deny your friend, but a ramp-check would find him flying without a valid medical.

BasicMed does not completely do away with SI's so your buddy may still need one if he has never had one and has one of the 4 qualifying issues.

To directly answer your question the insurance company may deny a claim on these grounds as they are somewhat free to deny claims as they see fit. There is some more info from the FAA related to SSRI's here. Remember that basic med is about considering yourself fit to fly and being self conscious of your current state it does not set any protections for you that once you deem yourself fit to fly you actual are. Its a fairly new rule so only time will tell how cases play out in the future relating to issues.

It doesn't matter what the drug is treating though, if the drug is on the DNI, he will need an SI. The OP's friend would fall under Q11-A (never had an SI) because as I understand the OP's question, he may be eligible for one, but doesn't have one.

@RonBeyer that is what I thought, I have reworded it to be a bit more clear. The OP also does not state if the person has ever had and los a medical which could be an issue as well.

Interesting, thanks for the feed back.  He had a third class in 2008 so within the look back period of Basic Med.  He started taking the drug around 2011.  He has never lost a medical.  He has never been medically deemed to have any of the 4 issues under Mental Health in Q11.   Thanks again.

OK let me ask like this.  If he had the SI totally legit all backed up.  Could insurance find a way to get out of paying the claim because of the drug even with the SI?  I understand this is all hypothetical.  I'm not asking you to speak for an insurance company.  I guess the better question is, How useful is renters insurance?  Do they look for any reason to get out of paying an indemnity?

@augerin there are a few to many variables there as that is pretty much completely situationally dependent, depends on the case, the insurer etc. I have renters insurance but cant say I have any experience making claims.

Understood.  Thanks again.  Probably his best bet to talk to a few insurance firms.  Insurance is easy to buy but not always easy to use.


text:
In an Airbus A320, when alpha protection is active, is pitch protection no longer available?
The A320 FCOM chapter 27 flight controls, normal law, protections, shows a diagram how the sidestick input is used when protections are active. When high angle of attack protection is active the side stick controls angle of attack (alpha protection to alpha max), so far so good.
In several places throughout the FCOM and that diagram I talked about Airbus states that the alpha protection has the highest priority.
Does this mean when alpha protection is active that pitch protection is no longer available? It makes sense that the nose of the aircraft is allowed to drop below the normally protected minimum of 15 degrees but is the maximum pitch of 30 degrees (or 25 depending on config and airspeed I think) still protected?
When alpha prot is active is it possible to pitch up to more than the maximum pitch limit? The diagram would suggest yes.

It may be they didn't bother detailing the behaviour because the aircraft won't have enough power to climb as steep and therefore will pitch down on the alpha protection alone anyway.

When you fly empty and fast (say 230kt+) and then pull up quickly you can pull 2.5 gs (load protection) and if you then manage to get the angle of attack high enough to trigger the alpha protection you could have the alpha protection active with enough excess energy to pitch to a high pitch angle. Pitch rate will be limited by alpha max then of course.... And at least in my flight simulator I can actually do that dynamically. I tested two commercially available flight sims for this, both got me to 30 degrees of pitch well before reaching actual low speed conditions where I run out of energy.

But yes, the alpha protection does also factor in the speed trend and pitch as far as this video suggests: https://www.youtube.com/watch?v=9fqy8uPzW90 (awesome video by the way, 9:12 for alpha protection)
So the alpha protection might force the nose down enough

Hi Jan. I tweaked the title of your question to try to better summarize what you're asking. Feel free to [Edit] further or to roll back the edit if you feel my edit changed your intent in any way.

Yes, it is active. I have seen and done high AOA exercises in the simulator and the green dashes (PFD markings) remain on the pitch limits even after the high angle of attack protection activates. That tells me as a pilot, the aircraft still has pitch attitude protections active. Here is a snapshot I took from a windshear recovery in an A320 (level D simulator) from YouTube with alpha floor activated. You can clearly see in the yellow box that the green dashes are still on. This gives the pilot an indication of the pitch attitude protection.

It makes a lot of sense. The protection adds thrust when it enters alpha floor. This means, it should not require a lot of nose down pitch attitude to recover. High angle of attack does not prevent you from a fully developed stall. It recovers way before you enter a full stall. So, you do not need to pitch down by a lot. Even if you pull the stick fully back, the aircraft will maintain V alphamax and slowly climb as TOGA thrust is added by the activation of alpha floor. Thus, a higher pitch attitude should not be required in an escape maneuver like in windshear. As a matter of fact, if for any reason (maybe due to an upset) if the aircraft pitches up more than 50 degrees or pitches down more than 30 degrees, the abnormal attitude law activates, which forces the aircraft into alternate law to help the pilot recover from what the airplane considers something beyond its protection envelope.
The next hint is found in the FCOM. It states:

Furthermore, there is no emergency situation that requires flying at excessive attitudes. For these
reasons, pitch attitude protection limits pitch attitude:
‐ 30 ° nose up in conf 0 to 3 (progressively reduced to 25 ° at low speed).
‐ 25 ° nose up in conf FULL (progressively reduced to 20 ° at low speed).
‐ 15 ° nose down (indicated by green symbols “=” on the PFD’s pitch scale).

It further says that:

Pitch attitude protection enhances high speed protection, high load factor protection, and high
AOA protection.

The last statement says it all, does not it? Airbus believes pitch attitude protection helps to enhance the effectiveness of other protections, including high AOA protection. Again, it is sensible. You need limitations for the protections to ensure that they do not go berserk.

Thank you for the long and detailed reply. What I am thinking about is a situation when you fly 300kt or so pull up with full aft stick so that you trigger the high AOA protection just as you pitch through about 25deg of pitch and then you keep maintaining full aft stick. It now commands alpha max and if you have enough speed remaining (much more than for 1g stall) it should enable you to pitch beyond 30deg nose up. The question is, if alpha prot is active, is the pitch protection active. The green bars are on but the diagrams show otherwise.

As mentioned in other comments it could be that in alpha prot the aircraft is not actually flying AOA but instead is a modified attitude hold that uses the speed trend, pitch and bank to define a desired max pitch angle and tried to fly that instead. In a demo video by airbus the pilot demonstrating the maneuver says that they are interested in stable attitude and speed. This means you can't actually pitch up even though physically it would be possible with enough speed.

A320 is fly by wire and controlled by the computers. Sometimes we tend to forget that. The computers will limit the pitch even if you have the energy to increase the pitch attitude. But yes it can increase pitch attitude in an upset situation (due to a disturbance). And if it increases by an amount the control laws degrade.

@Jan Which diagram are you talking about. I am not entirely sure which one it is. Thanks.

I just now added the diagram to my question.

In normal law « Alpha prot » has the highest priority, if AOA is less than alpha limit, teta (pitch angle) is limited between +25 and -15 at long run. At long run,  since in emergency descent you are allowed for a short time to go beyond - 15° pitch down, to do that you need to pitch down the side stick very quickly after having engaged a steep roll.
If AOA protection is active, I don’t see how you can exceed 25°or 30° pitch up, since the FBW will pitch down the aircraft to follow the AOA protection.

The fundamental question is: does the AOA protection control angle of attack or does it control pitch and speed (trend)? If it does control AOA and nothing else it it physically possible to fly at 30° nose up and be at an angle of attack where the alpha protection kicks in and with a good enough thrust to weight ratio and a high speed (barely enough to fly above alpha protection) the protection would just follow your alpha command and since there is excess energy it could keep following that AOA and pitch above 30°. Unless there is an extra control loop on pitch angle.

Alpha protection is the highest protection, but lower priority protections will be followed if not contradicting higher priority. So the pitch CANNOT BE MAINTAINED ABOVE 25° for more than few seconds


text:
Airbus A320 automatic speed mode protection in SRS vertical mode
According to the A320 FCOM, capter 22 auto flight, flight guidance, the "automatic speed mode protection" of the auto flight system of the A320 disengages the flight directors upon reaching current VMAX+4kt which will cause A/THR to revert to speed mode. But this protection is only active in climb modes: CLB, OP CLB and EXP CLB.
Does anybody know why the "automatic speed mode protection" is not available in SRS vertical mode where A/THR is also in THR CLB mode?
Also does anybody know if this protection uses the VFE of the selected flap config or of the actual config? E.g. if you are in climb and inadvertently select flaps from UP to CONF ONE then the aircraft won't let you do that but does this (cruise lock) but as far as I know the speed tape on the PFD already shows the maximum airspeed for the selected flap 1 config. So does this already disengage the flight directors? Or do they use the actual flap position or overspeed warning from the flight warning computers?

High speed protection is still available (in normal law) in SRS (check FCOM DSC 27 20-10-20).
What you will not have is the reversion mode of the autothrust to Speed. You will get the protection  (THS freeze, spiral stability to 0deg, max 40 deg bank, nose up order) but thrust will remain in THR CLB.
I suspect the reasoning was that SRS mode implies you are pretty low on altitude, and you do not desire an automatic (and uncommanded) reduction of thrust on, say, a windshear.
Of course, in this particular case the high speed prot will activate at VMO/MMO so if you manage to get that fast while still in SRS you have bigger problems than the thrust
For your second question: mode reversion to Speed will occur at VMAX (which is lower of VMO, VLE or VFE), but for SRS there is no flap overspeed protection so only VMO will trigger it

yes, but protection might be an unfortunate term here. if you ignore the FDs there is nothing to keep the V2+15.It is just the managed(target) speed that is maxed out to V2+15

SRS mode is a TAKE OFF or GO AROUND mode, the thrust is fixed and the pitch will maintain the min speed to V2 +10 with 2 engines and V2 in case of an engine failure; this mode requires slats being extended and it stops at acceleration altitude. Speed protection is provided by pitch attitude no need for other speed protection.
With respect to the second part of the question:
It is the SFCC active S/F position not the selected one that is used by APFD
In case of protection, I.e. divergence between the selection and the actual surfaces position, the autopilot flight director uses the actual surfaces position. The speeds prot indications that show on the PFD are computed by the FAC (flight augmentation computer) they do not affect the APFD neither any reversion directly.

The thrust is also fixed in OP CLB or CLB and the pilots could chose not to fly V2 + 10kts with SRS engaged, e.g. if they don't pitch up quick enough, so the question is what happens if you fly too fast during takeoff or go-around, e.g. with flaps 3 set and bust the speed limit. Sure nobody does that but the scenario is the same: pilot not pitching up and thrust very high.

You can’t fly too fast with flaps at position 3 unless you fly level, nevertheless no changes to the modes will occur before acceleration altitude, however note flap 3 is not a TAKE OFF valid position, so this scenario will trigger the take off config alarm, if by leveling before the acceleration altitude you keep increasing the speed, you will get the over speed alarm but no mode REVERSION

Conf 3 is a valid takeoff position

You are right, I was thinking of v2500


text:
the usage of thin airfoils for high-speed flight
what is The usage of thin aerofoils for high-speed flight
I tried those answers:
   They are perfect for controlling the local angle of attack, for instance in turbomachinery and flaps.
   At high speeds such as supersonic and transonic, because conventional airfoils cause wave drag.
   They are used at high-speed aircrafts such as Military aircraft.
but am not sure if they are correct!....or if they are correct someone can add more answers.

The third is circular reasoning. You're saying the usage of X for high-speed flight is that X is used for high-speed flight. (I'm not commenting on the correctness of the statement.)

This sounds like a homework question.

The wording of the question doesn't really make sense. What is the usage of thin airfoils for high speed flight? Well, strictly speaking, what their usage is... is defining the streamwise profile of the wing.
Based on the answers you've provided, I'll assume the question is actually asking:
Why are thin airfoils used for high speed flight?
Lets look at your 3 answers:
They are perfect for controlling the local angle of attack, for instance in turbomachinery and flaps.
Perfect is a strong word. Thin airfoils certainly could be well-suited for those applications, seeing as you'd want to avoid having really thick airfoils due to possible stall and blockage. I'm not sure what is meant by "Controlling the local angle of attack". In the case of both a flap and a compressor blade (turbomachinery), the airfoil doesn't determine what angle of attack the airfoil sees. That depends on how it's mounted or actuated. For a compressor blade, the angle of attack it sees would be influenced by the preceding stator. Perhaps the answer is trying to imply you can better control the flow angle coming off a thin airfoil than off a thick one? But the usage of angle of attack implies that somehow changing the airfoil will change the local angle of attack. This is untrue because the local angle of attack is defined between the chord line and the freestream. Simply changing the airfoil thickness while holding all else constant will not change the local angle of attack.
At high speeds such as supersonic and transonic, because conventional airfoils cause wave drag.
This is true. Wave drag rises more slowly for thin airfoils than thick ones. Onset of shock is delayed leading to lower wave drag.
They are used at high-speed aircrafts such as Military aircraft.
This is true, but it doesn't really answer your question (if I am interpreting it correctly). It is describing a use case for thin airfoils, not why they are used for high speed flight.
I had a very difficult time parsing your question because it's difficult to tell if It's asking about when thin airfoils are used or why they are used.